In a YDSE arrangement, the distance of screen from the slits is half the distance between the slits. If `'lambda'` is the wavelength of then the value of D such that first minima on the screen falls at a distance D from then centre 'O' is

Options

A. `D = (lambda)/(sqrt(5))`
B. `D = (lambda)/(2)`
C. `D = (lambda)/(5sqrt(2))`
D. `D = (lambda)/(2sqrt(5))`

Answer

D

Explanation

`T_(2)P = D+X, T_(1)P = D-X`
`S_(1)P = sqrt((S_(1)+T_(1))^(2)+(PT_(1))^(2))= [D^(2)+(D-X)^(2)]^(1//2)`
`S_(2)P= [D^(2)+(D-X)^(2)]^(1//2)`
Minima will occur when `S_(2)P- S_(1)P = (lambda)/(2)`
If `X = D , (5D^(2))^(1//2) = (lambda)/(2) , :. D = (lambda)/(2sqrt(5))`